Question

A closed flat loop conductor with radius 2mm is located in a changing uniform magnetic field. If the emf induced in the loop is 2 V what is the rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies

Answer 1

Answer:

Rate of magnetic field is 15923.668 T/sec

Explanation:

We have given radius of loop conductor r = 2 mm = 0.002 m

Induced emf = 2 volt

As the magnetic field and plane are perpendicular to each other so angle between magnetic field and area is 0°

Cross sectional area of the conductor is equal to $A=\pi r^2$

$A=3.14\times 0.002^2=1.256\times 10^{-6}m^2$

Induced emf is given by $e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}$

$2=1.256\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=159235.668T/sec$